Use Picard’s theorem to show that the initial value problem $(1+e^x)\frac{dy}{dx} = \sin(x + y^3)$, $y(1) = 3$, has a unique solution on the interval $x ≥ 1$.

By Picard's Existence and Uniqueness Theorem; If $f$ is continuous on a domain $D$ and $f$ satisfies Lipschitz condition on $D$. If $R=\{|x−x_0|≤a;\,|y−y_0|≤b\}$ lies in $D$ and $M={\rm maximum}|f(x,y)|$, $\alpha=\min\{a,b/M\}$. Then the IVP has a unique solution on the interval $|x−x_0|≤\alpha$.

I'm not sure how to find the value of M and alpha and then go on to find the interval.

I've figured out that M=1/(1+e^(1-a)) and alpha= min(a,b(1+e^(1-a)) however I cant figure out a and b.


For existence theorem to be used f(x,y) needs to be continuous in the interval \begin{equation} R=\left\{(x, y):\left|x-x_{0}\right| \leq a,\left|y-y_{0}\right| \leq b\right\}, \quad(a, b>0) \end{equation} To find the interval actually you should randomly pick the area by yourself because you are analysing the equation and you pick those values of a and b according to the function so, you should pick some interval and test it with the existence and uniqueness theorems but they are not fully telling you the interval of validity so you should find the interval first according to some techniques.

  • $\begingroup$ so if was to choose a = 1 and b =1 my answer would be 0≤x≤2 based on |x−x0|≤alpha but this is not consistent with x ≥ 1 which is what the question is looking for $\endgroup$
    – Broccoli
    Apr 4 '20 at 16:47
  • $\begingroup$ doesn't this mean i've got the question wrong $\endgroup$
    – Broccoli
    Apr 4 '20 at 16:52
  • $\begingroup$ @Martin R would you be able to help $\endgroup$
    – Broccoli
    Apr 4 '20 at 16:55
  • $\begingroup$ google.com/url?sa=t&source=web&rct=j&url=http://… you can refer this PDF to solve your question it has quite good explanation but for your question it is pretty hard to solve the DE apparently. $\endgroup$
    – asd.123
    Apr 4 '20 at 17:11
  • $\begingroup$ What I can tell you is that you can try some intervals for the solution but this will be partially your question because you want to find that the solution of differential equation is unique at the interval of validity but using existence and uniqueness theorem you can most probably reach partial result to show that your solution to DE is unique you should solve the DE then check that is unique. I guess it answers your question $\endgroup$
    – asd.123
    Apr 4 '20 at 17:20

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